Why Estimate Binding Rates From Time Courses?
An equilibrium measurement describes the fraction of receptors bound after the system has settled. It does not, by itself, describe how quickly binding and unbinding occur. Time-course measurements provide that kinetic information: association observations track occupancy as ligand is added, while washout observations track occupancy after free ligand is removed. Under a one-site reversible binding model, these data can estimate the association rate constant \(k_{\mathrm{on}}\) and dissociation rate constant \(k_{\mathrm{off}}\).
The ratio of these constants has a direct equilibrium interpretation. That link is useful, but it depends on the model and experimental conditions: the ligand concentration must be known and effectively constant during association, and washout must remove free ligand sufficiently to prevent rebinding. The estimates are only as meaningful as these assumptions.
Model and Assumptions
Let \(\theta(t)\) denote the fraction of receptor sites occupied at time \(t\), so \(0\leq\theta(t)\leq1\). Let \(L\) be the free-ligand concentration. For one reversible binding site, the rates of change in fractional occupancy are proportional to the unoccupied fraction and occupied fraction, respectively.
For washout, the relevant assumption is different: after free ligand is removed, its concentration is approximately zero and rebinding is negligible. Under that condition, the association term vanishes. If free ligand remains, or released ligand can rebind, the observed decline need not represent dissociation alone.
What the Association and Washout Curves Estimate
With constant \(L\), the association equation can be written as \(\frac{d\theta}{dt}=k_{\mathrm{on}}L-(k_{\mathrm{on}}L+k_{\mathrm{off}})\theta\). Its equilibrium occupancy is \(\theta_{\infty}=k_{\mathrm{on}}L/(k_{\mathrm{on}}L+k_{\mathrm{off}})\). Define the observed relaxation rate \(\lambda=k_{\mathrm{on}}L+k_{\mathrm{off}}\), with units of time\(^{-1}\). The solution from an initial occupancy \(\theta_0\) is
Thus an association time course can estimate both its limiting occupancy and its relaxation rate. If the system starts unoccupied, then \(\theta(t)=\theta_{\infty}(1-e^{-\lambda t})\). When the plateau and \(\lambda\) are estimated reliably at a known \(L\), both rate constants can be recovered.
During washout, \(L\approx0\), so the model becomes \(\frac{d\theta}{dt}=-k_{\mathrm{off}}\theta\). For washout beginning at time \(t_w\), with occupancy \(\theta_w\), the solution is
Taking logarithms gives a straight-line relationship between log occupancy and elapsed washout time. Its slope is \(-k_{\mathrm{off}}\), provided occupancy is positive and the single-exponential washout model is appropriate. Measurements near the assay’s lower limit can make this transformation unstable.
Estimating Rates From Measurements
For association data, a practical first estimate of \(\lambda\) can come from fitting the exponential curve, or from a linear fit to \(\ln|\theta_{\infty}-\theta(t)|\) against time when \(\theta_{\infty}\) is known. The slope of that transformed relationship is \(-\lambda\). Once \(\theta_{\infty}\), \(\lambda\), and \(L\) are available, rearrangement yields
These equations presume the association measurements resolve the plateau well enough to estimate \(\theta_{\infty}\), and that the one-exponential form describes the approach to it. If occupancy begins above the plateau, the absolute difference in the logarithmic transformation is used, and the curve approaches the plateau from above.
For washout data, fitting \(\ln\theta(t)\) against \(t-t_w\) gives a slope of \(-k_{\mathrm{off}}\). With only two measurements, the rate estimate is the negative log change divided by elapsed time. With several observations, a regression or direct nonlinear fit uses more of the data and can reveal departures from a single exponential. In either case, assay error and model adequacy matter; a numerical estimate alone does not verify the assumptions.
Worked Example: Estimating \(k_{\mathrm{off}}\) From Washout
In a hypothetical washout experiment, measured occupancy is \(0.800\) at the start of washout and \(0.241\) six minutes later. Assume free ligand was rapidly removed, rebinding is negligible, and a single exponential describes the decline. Estimate \(k_{\mathrm{off}}\).
Under these assumptions, \(\theta(t)=\theta_w e^{-k_{\mathrm{off}}t}\), so \(k_{\mathrm{off}}=[\ln(\theta_w)-\ln(\theta(t))]/t\). Substituting the measurements gives
The estimated dissociation rate constant is approximately \(0.2000\ \mathrm{min}^{-1}\). The arithmetic is also consistent with the direction of change: occupancy declines, so the fitted washout slope is negative and \(k_{\mathrm{off}}\), defined as a positive rate constant, is positive. This two-point estimate cannot show whether the full washout curve is truly single-exponential; additional time points are needed to assess that.
Worked Example: Estimating Both Rates From Association
Suppose a hypothetical association experiment starts with unoccupied receptors. The free-ligand concentration is held at \(10.0\ \mathrm{nM}\), and the measured curve is well described by an exponential approaching a plateau of \(\theta_{\infty}=0.750\). A fit gives \(\lambda=0.400\ \mathrm{min}^{-1}\). Assume the one-site model is appropriate and the plateau is adequately measured.
The rate estimates follow from the association equations:
The implied equilibrium dissociation constant is
As a consistency check, the equilibrium occupancy predicted at \(10.0\ \mathrm{nM}\) is \(10.0/(10.0+3.33)\approx0.750\), matching the fitted plateau. This agreement is expected because the rates were derived from that plateau and relaxation rate; it is not an independent validation of the model.
Worked Example: Checking Estimates Across Ligand Concentrations
Two hypothetical association experiments use the same receptor preparation but different constant free-ligand concentrations. At \(L=5.00\ \mathrm{nM}\), the estimated plateau is \(0.600\) and \(\lambda=0.250\ \mathrm{min}^{-1}\). At \(L=20.0\ \mathrm{nM}\), the estimated plateau is \(0.857\) and \(\lambda=0.700\ \mathrm{min}^{-1}\). Calculate the rates from each experiment and compare them.
At \(5.00\ \mathrm{nM}\),
At \(20.0\ \mathrm{nM}\), using the displayed plateau rounded to three significant figures,
The estimates agree to the precision of the rounded inputs. The corresponding ratio is approximately \(0.100/0.0300=3.33\ \mathrm{nM}\). Similar estimates across concentrations support, but do not prove, the assumed model. Systematic disagreement could indicate noisy or poorly resolved plateaus, concentration errors, or model misspecification.
Choosing a Fit and Checking Its Assumptions
Log-linear transformations are useful for understanding the rate information, but they change the error structure. If measurement errors are additive on the occupancy scale, taking logarithms generally does not leave errors with constant variance. The transformation can also give disproportionate influence to small occupancies and requires positive measurements. Directly fitting the exponential model to untransformed observations is often preferable when its measurement-error assumptions are appropriate. A joint fit across multiple ligand concentrations can estimate shared \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\) while allowing each curve to have its own known concentration and, when justified, initial occupancy.
The main alternatives answer related but distinct questions. A washout-only fit estimates \(k_{\mathrm{off}}\), but does not identify \(k_{\mathrm{on}}\) without further information. An association fit at one concentration can estimate both rates if its plateau and relaxation are well resolved and the model holds. An equilibrium-only measurement can estimate \(K_D\) from occupancy versus concentration, but does not separate the two kinetic rate constants. Use agreement between estimates from association and washout as a diagnostic when both types of data are available.
Check plots of observed and fitted occupancy over time, residuals against time, and whether estimates are stable when individual time points are removed. Curvature in a log-washout plot, a systematic residual pattern, or rate estimates that change substantially across ligand concentrations may indicate that a single exponential is inadequate. Possible causes include multiple receptor states, changing free-ligand concentration, rebinding during washout, or measurement artifacts. These patterns warrant investigation; they do not by themselves identify which alternative mechanism is responsible.
Common Mistakes and Limitations
- Calling the association slope \(k_{\mathrm{on}}\). The fitted relaxation rate is \(\lambda=k_{\mathrm{on}}L+k_{\mathrm{off}}\), not \(k_{\mathrm{on}}\) alone. It must be combined with the plateau and ligand concentration to separate the two rates.
- Ignoring units. \(k_{\mathrm{on}}\) has concentration\(^{-1}\) time\(^{-1}\) units; \(k_{\mathrm{off}}\) has time\(^{-1}\) units; \(K_D\) has concentration units. Their ratio \(k_{\mathrm{off}}/k_{\mathrm{on}}\) therefore has the units required of a concentration.
- Interpreting washout as pure dissociation when ligand remains. Residual free ligand can support rebinding. In that case, a simple exponential washout fit may not estimate the underlying \(k_{\mathrm{off}}\).
- Treating a fitted plateau as proof of equilibrium. A curve that appears flat over the observed interval may still not have reached its true limiting occupancy. Limited time coverage can confound the plateau and relaxation-rate estimates.
The model also assumes one class of binding sites and constant kinetic rates. If those assumptions fail, the fitted values may be effective summaries rather than uniquely interpretable microscopic rate constants. Rate estimates should therefore be reported with the experimental ligand concentration, units, fitting method, and evidence that the assumed time-course form is adequate.
Check Your Understanding
For each question, work from the stated model and assumptions. Solutions are provided below the questions.
- In a washout experiment, occupancy falls from \(0.900\) to \(0.450\) over \(5.00\) minutes. Assuming negligible rebinding and a single exponential, estimate \(k_{\mathrm{off}}\).
- An association experiment at \(L=4.00\ \mathrm{nM}\) gives \(\theta_{\infty}=0.500\) and \(\lambda=0.280\ \mathrm{min}^{-1}\). Estimate \(k_{\mathrm{on}}\), \(k_{\mathrm{off}}\), and \(K_D\).
- Why does a single fitted association relaxation rate not by itself equal \(k_{\mathrm{on}}\)?
- Name two conditions that should be checked before interpreting a washout slope as \(-k_{\mathrm{off}}\).
Solutions. (1) \(k_{\mathrm{off}}=[\ln(0.900)-\ln(0.450)]/(5.00\ \mathrm{min})=\ln(2)/5.00\approx0.1386\ \mathrm{min}^{-1}\). (2) \(k_{\mathrm{on}}=\lambda\theta_{\infty}/L=(0.280)(0.500)/4.00=0.0350\ \mathrm{nM}^{-1}\mathrm{min}^{-1}\); \(k_{\mathrm{off}}=\lambda(1-\theta_{\infty})=(0.280)(0.500)=0.140\ \mathrm{min}^{-1}\); \(K_D=0.140/0.0350=4.00\ \mathrm{nM}\). (3) The relaxation rate is \(\lambda=k_{\mathrm{on}}L+k_{\mathrm{off}}\), so both association and dissociation contribute; estimating \(k_{\mathrm{on}}\) requires additional information, including the plateau, ligand concentration, and the assumed model. (4) Free ligand should be removed sufficiently that its concentration is approximately zero, and rebinding should be negligible; the washout curve should also be consistent with a single exponential over the measured range.